Does Time Dilation Occur with Uniform Motion?

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Assume I am traveling in a spaceship at 0.9c towards a certain star, away from Earth. Then, some guy on Earth shines a really bright LASER light at the same star, causing a beam right beside my ship, so that the light and my ship travel in parallel paths. Because I am traveling at 0.9c, I should see that LASER to be going at 0.1c right? But, when I am moving, the time interval should be smaller (to get from one place to another). Say that The LASER took 1 second to travel 3x10^8 meters. To me, would it be going faster, since less than 1 second has passed for me while it traveled the same distance?

v = d/t

Since t is smaller because I'm moving at 0.9c, v becomes bigger. Would c be bigger in my moving frame of reference?
Also, does time slow down only when you accelerate or when you travel with uniform motion also?
 
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fred2028 said:
Assume I am traveling in a spaceship at 0.9c towards a certain star, away from Earth. Then, some guy on Earth shines a really bright LASER light at the same star, causing a beam right beside my ship, so that the light and my ship travel in parallel paths. Because I am traveling at 0.9c, I should see that LASER to be going at 0.1c right?
No, velocities don't add that way in relativity. You have to use the formula for relativistic velocity addition, which says that if A is traveling at speed v relative to B, and B is traveling at speed u relative to C, then A's speed relative to C is (v + u)/(1 + u*v/c^2). In this case, if A is the light beam and B is the Earth, so v=1c, then if C is your ship, so u=-0.9c (negative because in your frame the Earth is moving in the opposite direction as the light beam is moving in the Earth's frame), then you will measure the velocity of the light beam as (1c - 0.9c)/(1 - 1*0.9) = 0.1c/0.1 = 1c. It is in fact one of the fundamental postulates of relativity that each observer should measure light to move at the same speed in his own rest frame.
fred2028 said:
But, when I am moving, the time interval should be smaller (to get from one place to another). Say that The LASER took 1 second to travel 3x10^8 meters. To me, would it be going faster, since less than 1 second has passed for me while it traveled the same distance?

v = d/t

Since t is smaller because I'm moving at 0.9c, v becomes bigger. Would c be bigger in my moving frame of reference?
To understand why each observer measures a light beam to move at c, you must take into account the fact that each observer sees a moving observer's rulers to be shrunk, their clocks to be slowed down, and their clocks to be out-of-sync with one another. I gave an example showing how all these factors together ensure they both measure the light's speed to be c in post #6 from this thread.
fred2028 said:
Also, does time slow down only when you accelerate or when you travel with uniform motion also?
Keep in mind that there's no "objective" truth about whose clocks slow down--each observer measures the other one's clocks to run slower than their own. However, from the perspective of a given observer, a moving clock will always be slowed down regardless of whether it's accelerating or moving inertially; at any given instant, if its instantaneous velocity is v, its rate of ticking at that instant is \sqrt{1 - v^2/c^2}.
 
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...and always remember that when dealing with relativity and "modern physics", you sort of want to treat it like politics... take 90% of rational thinking and throw it out the window...

another suggestion...read the Mr. Tompkins book...
 
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dingpud said:
...and always remember that when dealing with relativity and "modern physics", you sort of want to treat it like politics... take 90% of rational thinking and throw it out the window...

another suggestion...read the Mr. Tompkins book...

Yes, read Mr. Tompkins in Wonderland, its a fun read.

HOWEVER, it is not rational thinking you need to throw out it is preconceived notions as to how you think it OUGHT to be. Rather then closing your eyes and dreaming, study the math and let the MATH guide you to the correct and rational conclusions.
 
yes you are right...

I must say, when I first began studying relativity, it was very odd... like you said, preconceptions on how things "are" from environmental influence vs. how they are after "proving them with math's" can be a totally different thing...

spacetime still gets me sometimes...
 
I started reading a National Geographic article related to the Big Bang. It starts these statements: Gazing up at the stars at night, it’s easy to imagine that space goes on forever. But cosmologists know that the universe actually has limits. First, their best models indicate that space and time had a beginning, a subatomic point called a singularity. This point of intense heat and density rapidly ballooned outward. My first reaction was that this is a layman's approximation to...
Thread 'Dirac's integral for the energy-momentum of the gravitational field'
See Dirac's brief treatment of the energy-momentum pseudo-tensor in the attached picture. Dirac is presumably integrating eq. (31.2) over the 4D "hypercylinder" defined by ##T_1 \le x^0 \le T_2## and ##\mathbf{|x|} \le R##, where ##R## is sufficiently large to include all the matter-energy fields in the system. Then \begin{align} 0 &= \int_V \left[ ({t_\mu}^\nu + T_\mu^\nu)\sqrt{-g}\, \right]_{,\nu} d^4 x = \int_{\partial V} ({t_\mu}^\nu + T_\mu^\nu)\sqrt{-g} \, dS_\nu \nonumber\\ &= \left(...
In Philippe G. Ciarlet's book 'An introduction to differential geometry', He gives the integrability conditions of the differential equations like this: $$ \partial_{i} F_{lj}=L^p_{ij} F_{lp},\,\,\,F_{ij}(x_0)=F^0_{ij}. $$ The integrability conditions for the existence of a global solution ##F_{lj}## is: $$ R^i_{jkl}\equiv\partial_k L^i_{jl}-\partial_l L^i_{jk}+L^h_{jl} L^i_{hk}-L^h_{jk} L^i_{hl}=0 $$ Then from the equation: $$\nabla_b e_a= \Gamma^c_{ab} e_c$$ Using cartesian basis ## e_I...

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